Question

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answer 1

Number of observations (n) = 100

Incorrect mean $(\bar{x})=20$

Incorrect standard deviation $(σ)=3$

$⇒ 20=\frac{1}{100}\sum_{i=1}^{100}=x_i$

$⇒\sum_{i=1}^{100}x_i=20×100=2000$

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000-21-21-18 = 2000-60 = 1940

∴ $Correct\,mean=\frac{Correct\,sum}{100-3}=\frac{1940}{97}=20$

$Standard\,\, deviation\,σ=√\frac{1}{n}\sum_{i=1}^nx_i^2-\frac{1}{n^2}(\sum_{i=1}^nx_i)^2=√\frac{1}{n}\sum_{i=1}^2-(\bar{x})^2$

$⇒3=√\frac{1}{100}incorrect\,\sum_{i=1}^{n}x_i^2-(20)^2$

⇒ $incorrect\,\sum x_i^2=100(9+400)=40900$

∴ $Correct\,\sum_{i=1}^{n}x_i^2=incorrect\,\sum_{i=1}^{n}x_i^2-(21)^2-(21)^2-(18)^2$

$=40900-441-441-324$

$=39694$

∴ $Correct \,standard \,deviation=\,√\frac{Correct\,\sum{x_i^2}}{n}-(Correct\,mean)^2$

$=√\frac{39694}{97}-(20)^2$

$=√409.216-400$

$=√9.216$

$=3.036$